Molecular Orbitals several atoms Any group of atoms

Molecular Orbitals several atoms Any group of atoms: • Molecules, • Molecular ions, • Fragments, • Supermolecules 1

Describing molecular properties as a whole Molecular orbital theory is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule. 2

The orbitalar approximation Molecular orbital is a function that describes the wave-like behavior of a single electron in a molecule. A polyelectronic wave-function is expressed in terms (as a product or a determinant) of MOs. The use of the term "orbital" was first used in English by Robert S. Mulliken in 1925 as the English translation of Schrödinger's use of the German word, 'Eigenfunktion'. Robert Sanderson Mulliken 1996 -1986 Nobel 1966 3

Generalizing the LCAO approach: A linear combination of atomic orbitals or LCAO Sir John Lennard-Jones 1894 -1954 Linus Carl Pauling 1901 -1994 Nobel 1962 It was introduced in 1929 by Lennard-Jones with the description of bonding in the diatomic molecules of the first main row of the periodic table, but had been used earlier by Pauling for H 2+. 4

Any calculation starts by inputs and provides outputs Input Definition of the system Total number of electrons Choice of state (ground state or excited state) Nature of atoms involved (Z) Atomic Orbitals Geometry to define Potential Output Energy, Molecular Orbitals Electronic density Spin, Magnetism Search for minimization of energy The calculated geometry is the minimum found by optimization 5

Structure and Reactivity How the energy of a system respond to a structural variation? Changing distance or angle. 6

Symmetry To describe a molecule is describing • Its geometry and its symmetry • Its energy MOs are eigenfunctions of the symmetry operator and of H For H 2, we have use symmetries without solving the Schrodinger equation, For a larger system, symmetry helps simplifying or analyzing. One common symmetry is the plane for planar molecules; this causes s and p separation. To be considered when some parameter varies (structure modification, structure optimization, reactivity) symmetry has to be preserved during the variation 7

Valence orbitals size of the orbitals Usually, the atomic orbitals that participate to the MOs are valence orbitals. These are the valence orbitals of the neutral atom. For a cation in its highest oxidation state, this might be the unoccupied shell whereas the size of the cation is given by the outermost occupied shell See the case of Li+ 8

Valence orbitals size of the orbitals, the Li+ example Size of the cation Li+ according to Slater? The last occupied shell is the 1 s Z=(3 -0. 31) <r> = 1. 5/2. 69 a 0 = 0. 558 a 0 = 0. 295 Ả This is small (The Pauling ionic radius is also small, 0. 60 Ả) The valence orbital 2 s is the one involved in forming bonds Z=(3 -2*0. 85) <r> = 5/1. 3 a 0 = 3. 846 a 0 = 2. 034 Ả Atomic radius (empirical): 1. 45 Ả Atomic radius (calculated): 1. 67 Ả Covalent radius (2008 values): 1. 28 Ả Covalent radius (empirical): 1. 34 Ả van der Waals radius: 1. 82 Ả The covalent radius for the 2 s orbital is such that the orbitals overlap; it is less than the <r> value 9

Method to build M. O. s • Determine the symmetry elements of the molecule • Make the list of the functions involved (valence atomic orbitals) • Classify them according to symmetry (build symmetry orbitals if necessary by mixing in a combination the set of orbitals related by symmetry) • Combine orbitals of the same symmetry (whose overlap is significant and whose energy levels differ by less than 10 e. V). 10

Valence orbitals of H 2 O y. Oz Molecular plane x. Oz symmetry plane C 2 z axis 2 s. O S S S sigma 2 px. O A S A pi 2 py. O S A A sigma 2 pz. O S S S sigma 1 s 1 S - - sigma 1 s 2 S - - sigma Symmetry relative to the molecular plane, sigma-pi separation 11

Valence orbitals of H 2 O y. Oz Molecular plane x. Oz symmetry plane C 2 z axis 2 s. O S S S 2 px. O A S A 2 py. O S A A 2 pz. O S S S 1 s 1 S - - 1 s 2 S - - Symmetry relative to x. Oz, necessitates building symmetry orbitals Build them ! What are their energy levels? 12

Valence orbitals of H 2 O y. Oz Molecular plane x. Oz symmetry plane C 2 z axis 2 s. O S S S A 1 2 px. O A S A B 1 2 py. O S A A B 2 2 pz. O S S S A 1 (1 s 1+1 s 2) S S S A 1 (1 s 1 -1 s 2) S A A B 2 Redundancy: Three groups; A 2 is not present 13

C 2 v E y. Oz molecular plane x. Oz symmetry plane C 2 axis A 1 1 1 A 2 1 -1 -1 1 B 1 1 1 -1 -1 B 2 1 -1 14

Symmetry orbitals (1 s 1+1 s 2) The H-H distance is 2*sin(105/2)*1. 09 Ả = 1. 73 Ả (1 s 1 -1 s 2) This is long relative to 0. 74 Ả ! The energy level of the symmetry orbital is the atomic level for H, -13. 6 e. V 15

Values from Extended Hückel 1962 (-e. V) 2 s H 13. 6 Li 5. 4 He 24. 25 Be 10 B 15. 2 C 21. 4 N 26 O 32. 3 F 40 2 p 3. 5 6 8. 5 11. 4 13. 4 14. 8 18. 1 Na Mg Al Si P S Cl 3 s 5. 1 9 12. 3 17. 3 18. 6 20 30 3 p 3. 4. 5 6. 5 9. 2 14 13. 3 15 1 s Roald Hoffmann Nobel 1981 William Nunn Lipscomb, Jr. American 1919 Nobel 1976 16

1 Valence orbital of H 2 O Symmetry B 1 2 s. O A 1 2 px. O B 1 2 py. O B 2 2 pz. O A 1 (1 s 1+1 s 2) A 1 The atomic orbital 2 px. O is a lone in its group, It is a molecular orbital It has p symmetry. (1 s 1 -1 s 2) B 2 E 2 p(O)= -14. 8 e. V 17

2 Valence orbitals of H 2 O Symmetry B 2 2 s. O A 1 2 px. O B 1 2 py. O B 2 2 pz. O A 1 (1 s 1+1 s 2) A 1 (1 s 1 -1 s 2) B 2 E 2 p(O)= -14. 8 e. V and E 1 S(H) = -13. 6 e. V 18

2 s. O A 1 2 px. O B 1 2 py. O B 2 2 pz. O A 1 (1 s 1+1 s 2) A 1 (1 s 1 -1 s 2) B 2 3 Valence orbitals of H 2 O Symmetry A 1 E 2 s(O)= -32. 3 e. V E 2 p(O)= -14. 8 e. V And E 1 S(H) = -13. 6 e. V E 2 s is low in energy an d does not mix 19

6 AO → 6 MOs 3 A 1 → 3 A 1 2 B 2 → 2 B 2 1 B 1 → 1 B 1 What is the ordering of the levels? 20

Overlap involved in B 2 Overlap involved in A 1 At 45° same overlap angle HOH/2 = 105/2 = 52° 5 > 45° Splitting in B 2 > Splitting in A 1 21

2 B 2 6 AO → 6 MOs 3 A 1 → 3 A 1 4 A 1 2 B 2 → 2 B 2 1 B 1 → 1 B 1 In the Lewis formula, two electron pairs, two bonds The electron pairs are 2 s and 2 p Spectroscopy shows different electron levels The bonds are the result of two contributing MO A 1 and B 2 Orbital numbering: 1 A 1 for the core? 3 A 1 1 B 2 2 A 1 22

Hybridization MOs sp 3 agreement with VSEPR and Total density sp 2 closer to MOs and spectroscopy: two different pairs 23

The Walsh diagram 24

Reduction of symmetry 25

Typical Bond Angles in AH 2 Molecules molecule electronic configuration H-A-H bond angle Be. H 2 (1 a 1)2(1 b 2)2 180° BH 2 (1 a 1)2(1 b 2)2(2 a 1)1 127° CH 2 (1 a 1)2(1 b 2)2(2 a 1)1(b 1)1 134° CH 2 (1 a 1)2(1 b 2)2(2 a 1)2 102° NH 2 OH 2 Mg. H 2 Al. H 2 Si. H 2 PH 2 SH 2 1 a 1= 1 sg ; 1 b 2= 1 su (1 a 1)2(1 b 2)2(2 a 1)2(b 1)1 (1 a 1)2(1 b 2)2(2 a 1)2(b 1)2 (1 a 1)2(1 b 2)2(2 a 1)1 (1 a 1)2(1 b 2)2(2 a 1) 1(b 1)1 (1 a 1)2(1 b 2)2(2 a 1)2(b 1)2 103° 104° 180° 119° 118° 93° 92° 26

F. Walsh diagram for H 2 S 27

This is called an orbital correlation diagram: 28

C 2 s. C A 1 2 px. C B 1 2 py. C B 2 2 pz. C A 1 (1 s 1+1 s 2) A 1 (1 s 1 -1 s 2) B 2 3 Valence orbitals of CH 2 Symmetry A 1 E 2 s(C)= -21. 4 e. V E 2 p(C)= -11. 4 e. V And E 1 S(H) = -13. 6 e. V E 2 s. C – E 1 s. H < 10 e. V Hybrid orbitals Not atomic eigenfunctions A But part of the MOs eigenfunctions for the molecule 2 s-2 p. Z 2 s+2 p. Z 2 A 1 is bonding 3 A 1 is non-bonding 4 A 1 is antibonding 29

C 2 s. C A 1 2 px. C B 1 2 py. C B 2 2 pz. C A 1 (1 s 1+1 s 2) A 1 (1 s 1 -1 s 2) B 2 3 Valence orbitals of CH 2 Symmetry A 1 E 2 s(C)= -21. 4 e. V E 2 p(C)= -11. 4 e. V And E 1 S(H) = -13. 6 e. V E 2 s. C – E 1 s. H < 10 e. V Among the 3 A 1 valence MO, the lowest one is bonding and the middle one non-bonding; this differs from H 2 O! A 30

hydridization Justification: • s and p. Z both are of same symmetry and appear in the same linear combinations (MOs); hybridization allows anticipating. • hybridization aims combining AOs, each hybrid maximizing the interaction with a partner (here the symmetry orbital on Hs) and minimizing those with others. The idea is that we can thus separate the interactions with different partners. If correct, this is useful and allows transferability (replacing a substituent by another one involving the same hybrid). • Mathematically a set of hybrids is equivalent to a set of canonic orbitals Failure: • It is not possible to rigorously separate the interactions with different partners. Hybrids interact with each other. • Existing symmetries reappear if interactions between hybrids is included. 31

Methane The MO orbital description conflicts with sp 3 hybridization 32

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sp 3 hybridization 2 s 2 px 2 py 2 pz t 1 1/2 1/2 t 2 1/2 -1/2 t 3 1/2 -1/2 t 4 1/2 -1/2 <ti. Itj>= δij <ti. IHItj>= (E 2 s-E 2 p)/4 Eti = <ti. IHIti> = (E 2 s+3 E 2 p)/4 34

sp 3 hybridization x b b b b x b (E 2 s+3 E 2 p)/4 b b b x = 0 With b =(E 2 s-E 2 p)/4 E 2 p 1(E 2 s-E 2 p)/4 3 (E 2 s-E 2 p)/4 E 2 s 35

sp 3 hybrids along C 3 ¼ of 2 s 3/4 of 2 p 2 s 2 px 2 py 2 pz t 1 1/2 0 0 √ /2 t 2 1/2 √(2/3) 0 -1/√ 12 t 3 1/2 -1/√ 6 t 4 1/2 -1/√ 6 -1/√ 2 -1/√ 12 36

• • • Interest of hybridization Pictorial Anticipating AO combinations within MOs Global density Transferable: Analysis through substituents Writing VB structures What is bad with hybridization? • Not describing symmetry • Not good for spectroscopy • Lack of orthogonality between hybrid orbitals belonging to the same center (tails of localized orbitals) 37

sp 3 hybrids along C 2 2 s 2 px 2 py 2 pz t 1 1/2 1/√ 2 0 1/2 t 2 1/2 -1/√ 2 0 1/2 t 3 1/2 0 1/√ 2 -1/2 t 4 1/2 0 -1/√ 2 -1/2 38

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hybridization Mixing 2 s and 2 p: requires degeneracy to maintain eigenfunctions of AOs. Otherwise, the hybrid orbital is an average value for the atom, not an exact solution. This makes sense when ligands impose directionality: guess of the mixing occurring in 41 OMs.

Angular dependence Expression along C 2 (z is the main axis) t = a s + √(1 -a 2) [pzcosq ± px sinq] t = a’s + √(1 -a’ 2) [pzcosq’ ± py sinq’] 2 a 2+2 a’ 2=1 √(1 -a 2) sinq = 1/√ 2 √(1 -a’ 2) sinq’ = 1/√ 2 2 q 2 q’ : : q= 105. 5/2 (<109° 28) = 52. 75° → a = 0. 459 a’ = √(1/2 -a 2) = 0. 548 sinq’ = (1/√ 2)/(1/ √(1 -a’ 2)=0. 8453 q’ = 57° = 114. 7/2 > 109° 28 42

Angular dependence 2 q 2 q’ : : a 2+a’ 2= ½ √(1 -a 2) sinq = 1/√ 2 √(1 -a’ 2) sinq’ = 1/√ 2 (1 -a 2) sinq 2 = 1/2 =(1 -a’ 2) sinq’ 2 1 -1/2 sinq 2 = a 2 1 -1/2 sinq’ 2 = a’ 2 1 -1/2 sinq 2 +1 -1/2 sinq’ 2 = a 2 + a’ 2 2 = 1/2 sinq 2 +1/2 sinq’ 2 + ½ 3 = 1/sinq 2 +1/sinq’ 2 When q decreases q’ increases! 43

Weight of s orbital 2 q = 105° 5 water a is smaller than ½ a’ is larger than ½ The weight of s is larger in the lone pairs. The s level is lower in energy than the p level. It has to participate to the stabilization of lone pairs. L-, CH 3+ 90° pure p sp 3 ligand 109° 28’ ¼ s ¾ p CH 3 -, L+ >109° 28’ >1/4 s 44

For the reconstructed Silicon(100) surface, the s character is higher for the dangling bonds of the outmost atoms. - + 45

H 3 The orbitals of H 3 (equilateral triangle in the xy plane) Are symmetry orbitals of NH 3 and CH 3 matching p orbitals 1/√ 2 -b 2 px 2 py 2 b 2 pz 1/√ 3 46

H 3 1/√ 2 -1/√ 2 The orbitals of H 3 (equilateral triangle in the xy plane) are symmetry orbitals of NH 3 and CH 3 matching p orbitals -b 2 px 2 py 2 b 2 pz 1/√ 3 47

H 3 1/√ 2 -1/√ 2 Pseudo symmetry: for H 3 the OM are degenerate. They remain degenerate interacting 2/3 = 1/√ 2 + c 2 with the 2 p orbitals (E) -b 2 px √(2/3) 2 py Normalization 1 = (2/√ 3)2 +2 c 2 48

H 3 1/√ 2 -1/√ 6 The orbitals of H 3 (equilateral triangle in the xy plane) are symmetry orbitals of NH 3 and CH 3 matching p orbitals -b √(2/3) 2 px 2 py 2 b 2 pz 1/√ 3 49

NH 3 s system Lone pair Bonding 50

NH 3 1/√ 2 p system -1/√ 2 -1/√ 6 2 px 2 py √(2/3) 51

Ammonia 52

Free rotation of a methyl group adjacent to a p system Hyperconjugation CH 3 always has one orbital conjugating with the p system 53 Mirror symmetry pseudosymmetry

Free rotation of a methyl group X Z X x x p y s s y Z p 54

Orientation of CH 2 D conjugated with a CH 2+ D + or + D 55

D + + A D x CH 2 With a substituent, the 2 orbitals are not equivalent. That of highest energy level interacts more strongly with the CH 2 group 56

Ethane 57

Ethene (ethylene) 58

Ethyne (acetylene) 59

BF 3 60

Octet rule, eighteen electron rule N in NH 3 4 atomic orbitals 4 bonding orbitals or non bonding accommodating 8 electrons 3 antibonding orbitals C in CH 4 4 atomic orbitals 4 bonding orbitals accommodating 8 electrons 4 antibonding orbitals W in W(CO)6 9 atomic orbitals 9 bonding orbitals or non bonding accommodating 8 electrons 6 antibonding orbitals 61

Ligand field for octahedral environment W in W(CO)6 9 atomic orbitals 9 bonding orbitals or non bonding accommodating 8 electrons 6 antibonding orbitals The s interaction raises the d. X 2 and d. X 2 -d. Y 2 levels The p with p *CO stabilizes the “non bonding levels” 62

Ligand field for octahedral environment W in W(CO)6 9 atomic orbitals 9 bonding orbitals or non bonding accommodating 8 electrons 6 antibonding orbitals The s interaction raises the d. X 2 and d. X 2 -d. Y 2 levels The p with p *CO stabilizes the “non bonding levels” 63

Ligand field for octahedral environment Ti in Ti(H 2 O)2(HO)4 Or Ti in Ti. O 2 Ti in 9 atomic orbitals 9 bonding orbitals or non bonding accommodating 8 electrons 6 antibonding orbitals The s interaction raises the d. X 2 and d. X 2 -d. Y 2 levels The p interaction with p. O (oxygen pairs) is a stabilizing interaction. The same occurs with PH 3 ligands Rutile Ti. O 2 These 6 electrons count for the eighteen electron 64 rule

Ligand field for various environments 65

Ni(PH 3)2. . . CH 2=CH 2 nickel (Z= 28) Ni. D 2 where D is a donor substituent (group PH 3 ). Ni will be only represented by the set of d (x 2 -y 2, z 2, xy, xz et yz) orbitals taking E(3 d) = - 12 e. V. D is modeled by an s orbital with 2 electrons: E(D)= - 15 e. V. What is the best orientation for the C=C bond? 1) Draw an energy diagram for Ni. D 2 2) Why the ethylene molecule is represented above by its p* orbital? 3) Draw an energy diagram for the interaction and tell which orientation is the best. 66

Ni(PH 3)2. . . CH 2=CH 2 Interaction with dx 2 -y 2 Largest interaction Interaction with d(x+y), z weaker 67

z y * Formaldehyde CH 2 O O 68

B 1 s orbital With p overlap 69